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X eafiiest sueta for that Assume set. finite a be i g Y Introduction i → Contents ( aifigthat satisfying ,ϕ x, X 97,3B2 88,26A30. 28A80, 39B12, 39B72, ( 1 and x )) x , g i ∈ : X X i × i , Y ∈ → I. osy hsmeans This eously. dracertain a ider o harmonic not Y einvariant he n.I our In ons. ntoson unctions haseries a th w differ- two uhthat such r ie.Now given. are qain (cf. equations l i ufficient robabil- econsider we e ∈ Let . sfunctional ’s equations, e I subset a , X (1.1) and ated 27 26 23 19 6 3 1 2 KAZUKI OKAMURA case that X = [0, 1] and I = 0, 1 and gi : Y Y . De Rham’s functional equation driven by affine functions{ } and related→ functions such as Takagi functions have been considered in many papers. A few of related results are Hata [Ha85], Zdun [Z01], Girgensohn-Kairies-Zhang [GKZ06], Serpa- Buescu [SB15a, SB15b, SB15c], Shi-Yilei [ST16], Barany-Kiss-Kolossvary [BKK18], Allaart [A], etc. Here we do not give a detailed review of this topic. Recently, Serpa-Buescu [SB17] considered (1.1) and gave necessary conditions for existence of the solution of (1.1). In the case that X and Y are metric spaces, they also gave sufficient conditions for existence and uniqueness. They also gave explicit formulae for the solution. This paper has three purposes. The first one is to consider sufficient conditions for existence and uniqueness of the solution of (1.1). The second one is investigating regularity properties for the solution. The final one is considering a kind of stability of the solution. This paper is organized as follows. In Section 2, we consider two different series of sufficient conditions for existence of the solution of (1.1), which are stated in Theorems 2.5 and 2.9. These results are similar to [SB17, Theorem 1], however, in Theorem 2.5, we remove several assumptions of [SB17, Theorem 1], and furthermore, in Theorem 2.9 we deal with the case that fi is not injective. In Section 3, we consider local regularity of solutions via invariant mea- sures of iterated function systems. We focus on the case that for each i I, ∈ Xi = X and the value of gi(x,y) does not depend on x and furthermore (X, fi i∈I ) and (Y, gi i∈I ) are iterated function systems satisfying the open{ set} conditions. Informally{ } speaking, we state in Theorem 3.9 that un- der certain conditions the solution of (1.1) is fractal, or in another phrase, singular. The solution of (1.1) measures how “far” two iterated function systems (X, fi i∈I ) and (Y, gi i∈I ) are. Although we do not need to in- troduce measures{ } for the definition{ } of (1.1), we state Theorem 3.9 by using integrals of certain functions with respect to the invariant measures of iter- ated function systems (X, fi i∈I ) and (Y, gi i∈I ) equipped with a common probability weight. We emphasize{ } that Theorem{ } 3.9 is applicable to the case that fi and gi are non-affine functions. Theorem 3.9 generalizes a modified statement of [O16, Theorem 1] and is also related to [Ha85, Theorems 7.3 and 7.5] and [Z01, Theorems 6 and 7]. [Ha85, Z01, O16] deal with the case that X = [0, 1], however, our result is also applicable to the case that X is not [0, 1]. We deal with the case that X is the two-dimensional standard Sierpi´nski gasket. In Sections 4 and 5, we give several examples. In Proposition 4.2, we consider the case that fi is not injective by applying Theorem 2.9. In Ex- ample 4.11, we give an example for the case that infinitely many solutions exist. Example 4.7 deals with the case that the iterated function systems (X, fi i∈I ) and (Y, gi i∈I ) have overlaps. In Example 5.8, we give an ex- ample{ for} the case that{ }X is the two-dimensional standard Sierpi´nski gasket and Y = [0, 1]. The solution is different from fractal interpolation functions on the Sierpi´nski gasket considered by Celik-Kocak-Ozdemir [CKO08], Ruan [R10] and Ri-Ruan [RR11]. The solution is not a harmonic function on the 3 two-dimensional Sierpi´nski gasket, and we would be able to call the solution a “fractal function on a fractal”. In Section 6, we consider a certain kind of stability of the solution. A little more specifically, we justify the following intuition: If two systems of (1.1) are “close” to each other, then, the two solutions of these systems are “close” to each other. Since we do not put any algebraic structures on X or Y , we cannot consider the Hyers-Ulam stability. We consider an alternative candidate of stability, by using the notion of Gromov-Hausdorff convergence on the class of metric spaces. Finally in Section 7, we state four open problems concerning conjugate equations.
2. Existence and uniqueness Let I be a finite set containing at least two distinct points. Let I = 0, 1,...,N 1 . Let X be non-empty sets and (Y, dY ) be a complete {metric space.− Assume} that for each i I a map g : X Y Y is given. ∈ i i × → Assumption 2.1. For each i I, let Xi X and fi : Xi X be a map such that ∈ ⊂ → X = fi(Xi). i[∈I Henceforth, if we do not refer to Xi, then, we always assume that Xi = X. Let A be the set of contact points, that is, A := x X f (x )= f (x ) = f −1(f (X )), i { i ∈ i | i i j j } i j j j∈[I\{i} xj[∈Xj j∈[I\{i} −1 A := Ai = fi (fj(Xj)). i[∈I i[6=j Let A := A f f (A). ∪ i1 ◦···◦ in n[≥1 i1,...,i[n∈I Assumption 2.2.eAssume that there exists a unique bounded map ϕ0 : A Y such that (i)→ gi(xi, ϕ0(xi)) = gj(xj, ϕ0(xj)) holds for every xi Ai and xj Aj satisfying that fi(xi)= fj(xj). (ii) If f f ∈(x) A, then,∈ i1 ◦···◦ in ∈ ϕ (f f (x)) = g (f f (x), ) g (x, ) ϕ (x). 0 i1 ◦···◦ in i1 i2 ◦···◦ in · ◦···◦ in · ◦ 0 We say that a function ψ on R is increasing (resp. decreasing) if ψ(t1) ψ(t ) (resp. ψ(t ) ψ(t )) whenever t t , and, is strictly increasing≤ 2 1 ≥ 2 1 ≤ 2 (resp. strictly decreasing) if ψ(t1) < ψ(t2) (resp. ψ(t1) > ψ(t2)) whenever t1
We say that T is ψ-contractive in the sense of Browder (See Jachymski [J97] for details) if we can take ψ in the above as a strictly increasing function. Hereafter, if we say that a map is a weak contraction in the sense of Matkowski or Browder, then, we mean that for some ψ the map is ψ- contractive in the sense of Matkowski or Browder, respectively. We remark that if (2.1) holds for an increasing function ψ, then, ψ(t)
Assumption 2.4. Assume that each fi is injective and f −1(A) A. (2.2) i ⊂ i[∈I e e (2.2) is satisfied if x A whenever fi(x) A for some i I, or A = , for example. ∈ ∈ ∈ ∅ Theorem 2.5. Under Assumptions 2.1, 2.2, 2.3 and 2.4, there exists a unique bounded map ϕ : X Y such that ϕ = ϕ on A, and (1.1) holds for → 0 every i I and x Xi. Here the boundedness of ϕ means that Image(ϕ) is contained∈ in a ball∈ on Y , specifically,
sup dY (ϕ(x1), ϕ(x2)) < + . x1,x2∈X ∞ Proof. Let ϕ be a map on A defined by ϕ (y) := ϕ (y) if y A, and, 1 1 0 ∈ ϕ1 (fi fin (x)) := gi (fi fin (x), ) gin (x, ) ϕ0(x), x A, n 1. 1 ◦···◦ 1 e 2 ◦···◦ · ◦· · ·◦ · ◦ ∈ ≥ This is well-defined due to Assumption 2.2. We now check this. Assume for n,m 1 and x,y A, ≥ ∈ f f (x)= f f (y). i1 ◦···◦ in j1 ◦···◦ jm then, fi2 fin (x) A and fj2 fjm (x) A. Now use Assumption 2.2 (i). Assume◦···◦ for n ∈ 1,m = 0 and◦···◦x,y A, ∈ ≥ ∈ f f (x)= y A. i1 ◦···◦ in ∈ then, use Assumption 2.2 (ii). Recall Assumption 2.4. If x A A, then, f (x) A. By the definition ∈ i ∩ i ∈ of ϕ and Assumption 2.2, (1.1) holds for each i I and x A. 1 ∈ ∈ Hence, if X = A, then we have (1.1)e for each i I ande x X . ∈ ∈ i Assume X A = . Let be the set of bounded maps fromeX A to Y . \ 6 ∅ B \ By using Assumptione 2.4 and the definition of A, for each x X A, there e −1 ∈ \ e is a unique i I such that x fi(Xi) and fi (x) X A. Now we can∈ define T : ∈ such that e ∈ \ e B→B T [ϕ](x) := g f −1(x), ϕ(f −1(x)) , x X eA, i i i ∈ \ e 5 for every ϕ . We now put∈B the following metric on . B D(ϕ1, ϕ2):= sup dY (ϕ1(x), ϕ2(x)) < + , ϕ1, ϕ2 . e x∈X\A ∞ ∈B Then, ( , D) is a complete metric space. B Lemma 2.6. (2.1) holds for maxi∈I φi.
Proof of Lemma. For each t and n, there exists a sequence (ik)1≤k≤n such that n
max φi (t)= φi1 φin (t). i∈I ◦···◦ Then, by using the fact that φi(t) 2.2. Result for the case that gi does not depend on x. In the proof of Theorem 2.5, the injectivity of fi is needed. Now we investigate the case that the injectivity of fi fails. Assumption 2.8. Assume g = g (x, ). Let K be a unique compact subset i i · of Y such that K = i∈I gi(K). We assume that for any x A, there exists a unique infinite sequence∪ (i ) IN such that ∈ n n ∈ x Image(f f ), (2.4) ∈ i1 ◦···◦ in n\≥1 and furthermore, g g (K)= ϕ (x) . i1 ◦···◦ in { 0 } n\≥1 6 KAZUKI OKAMURA Theorem 2.9. Under Assumptions 2.1, 2.2, 2.3, and 2.8, three exists a unique bounded solution ϕ of (1.1) such that ϕ = ϕ0 on A. Remark 2.10. If gi = gi(x, ), then, this is an extension of Theorem 2.5. In the following proof, we do· not use any fixed point theorems. Proof of Theorem 2.9. First we show the existence. Let x X. We first remark that by Assumption 2.1, for any x X∈ there exists at N ∈ least one infinite sequence (in)n I satisfying (2.4). ∈ N If there exists a unique infinite sequence (in)n I satisfying (2.4), then, by Assumption 2.3, we let ϕ(x) Y be an element∈ such that ∈ g g (K)= ϕ(x) . (2.5) i1 ◦···◦ in { } n\≥1 Otherwise, there exists a maximal integer n = N(x) such that there exists a unique (i1,...,in) such that x Image(fi1 fin ). Then there exist at least two candidates of i ∈I and x ◦···◦A such that N(x)+1 ∈ N(x)+1 ∈ x = f f (x ) (2.6) i1 ◦···◦ iN(x)+1 N(x)+1 and let ϕ(x) := g g (ϕ (x )). i1 ◦···◦ iN(x)+1 0 N(x)+1 This is well-defined, that is, ϕ(x) does not depend on the choice of i N(x)+1 ∈ I and x A satisfying (2.6), due to Assumption 2.2. N(x)+1 ∈ We need to show that ϕ = ϕ0 on A. Let x A. By Assumption 2.8, there exists a unique infinite sequence (i ) ∈IN satisfying (2.4). Then n n ∈ (2.5) holds. Now by Assumption 2.8, ϕ(x)= ϕ0(x). Second we show the uniqueness. Let ϕ1 and ϕ2 be bounded the solution of (1.1). Let x X. If there exists∈ a unique infinite sequence (i ) IN satisfying (2.4), then, n n ∈ by (1.1) and the boundedness of ϕ1 and ϕ2, g g (Image(ϕ ) Image(ϕ )) = ϕ (x) = ϕ (x) . i1 ◦···◦ in 1 ∪ 2 { 1 } { 2 } n\≥1 Hence ϕ1(x)= ϕ2(x). Otherwise, there exists a maximal integer n = N(x) such that there exists a unique (i1,...,in) satisfying (2.4). Then there exists at least two candidates of iN(x)+1 I and xN(x)+1 A satisfying (2.6). By this, (1.1) and Assumption 2.2, ∈ ∈ ϕ (x)= g g (ϕ (x )). i i1 ◦···◦ iN(x)+1 i N(x)+1 By the uniqueness for ϕ0 in Assumption 2.2, ϕ0 = ϕ1 = ϕ2 on A. Hence ϕ1(x)= ϕ2(x). Remark 2.11. We are not sure whether there exist relationships between Assumptions 2.4 and 2.8. 3. Regularity In this section we always assume that (X, dX ) and (Y, dY ) are two compact metric spaces and that there exist weak contractions f , i I, on X and i ∈ 7 g , i I, on Y in the sense of Browder such that i ∈ X = fi(X) and Y = gi(Y ). i[∈I i[∈I Furthermore assume that there exists a unique solution ϕ of (1.1). The aim of this section is to give sufficient conditions for each of the following: Definition 3.1. Let α> 0 and a [0, + ]. (1) For a non-empty subset U of X∈, we say∞ that (α, U, a) holds if dY (ϕ(x1), ϕ(x2)) sup α = a. x1,x2∈U, x16=x2 dX (x1,x2) (2) For x X, we say that (α, x, a) holds if ∈ dY (ϕ(z), ϕ(x)) lim sup α = a. z→x dX (z,x) For a subset A of a metric space and for s,δ > 0, let ∞ ∞ δ s s(A) := inf diam(Ui) A Ui, diam(Ui) δ, i , H ( | ⊂ ≤ ∀ ) Xi=1 i[=1 where we let diam(U) be the supremum of distances of two points of U. This value is monotone decreasing with respect to δ, we can let δ s(A) := lim s(A). H δ→0+ H Then, we define the Hausdorff dimension of A as follows: dim (A) := sup s> 0 : (A)=+ = inf s> 0 : (A) = 0 . H { Hs ∞} { Hs } Lemma 3.2. Let (X, dX ) and (Y, dY ) be metric spaces. Let A X and B Y be non-empty. Let ϕ : A B be a surjective map.⊂ If α > dim⊂ (A)/ dim (B), then, (α, A, + )→holds. H H ∞ Proof. Assume that (α, A, + ) fails. Then, ∞ dY (ϕ(x1), ϕ(x2)) sup α < + , x1,x2∈A, x16=x2 dX (x1,x2) ∞ that is, ϕ is α-H¨older continuous. Therefore, in the same manner as in the proof of [Fal14, Proposition 3.3]1, we can show that dim (A) dim (ϕ(A)) H . H ≤ α Since ϕ is surjective, dim (A) dim (B) H . H ≤ α This contradicts the assumption that α> dimH (A)/ dimH (B). For a Borel probability measure µ on a metric space, let dim µ := inf dim K K : Borel measurable µ(K) > 0 . H { H | } Let (X, dX ) and (Y, dY ) be two compact metric spaces. Assume that there exist weak contractions f , i I, on X and g , i I, on Y in the sense of i ∈ i ∈ 1It is stated for subsets of the Euclid space, but it holds also for general metric spaces. 8 KAZUKI OKAMURA Browder such that X = fi(X) and Y = gi(Y ). i[∈I i[∈I For pi, i I, be numbers in (0, 1) such that i∈I pi = 1, let µ{pi} and ν be two∈ probability measures on X and Y such that {pi} P µ = p µ f −1, and ν = p ν g−1. {pi} i {pi} ◦ i {pi} i {pi} ◦ i Xi∈I Xi∈I The existences and uniquenesses of µ{pi} and ν{pi} under the assumption of the above theorem are assured by Fan [Fan96]2. Proposition 3.3. Assume that ϕ is a solution of (1.1). Assume dimH ν{pi} > 0. Let dimH µ{p } α> i . dimH ν{pi} Then, (α, U, + ) holds for every non-empty open set U of X. ∞ This assertion is useful if we can know the values of dimH µ{pi} and dimH ν{pi}. Fan-Lau [FL99] might be useful under certain regularity as- sumptions for the two IFSs (X, f ) and (Y, g ). { i}i { i}i Proof. Let α> dimH µ{pi}/ dimH ν{pi}. Let U be an arbitrarily open set of X. Take ǫ> 0 such that dimH µ{p } + ǫ α> i . dimH ν{pi} Then, we can take A X such that ⊂ dim A dim µ + ǫ. H ≤ H {pi} Since X is compact and fi are weak contractions, there exists i1,...,in such that f f (A) U. i1 ◦···◦ in ⊂ Since each fi is Lipschitz continuous, dim f f (A) dim A. H i1 ◦···◦ in ≤ H Take a Borel subset B of Y arbitrarily. Then, by the definition of µ{pi} −1 −1 −1 µ{pi}(ϕ (B)) = piµ{pi} fi (ϕ (B)) . i∈I X By (1.1), we have that for each i I, ∈ f −1(ϕ−1(B)) = (ϕ f )−1(B) = (g ϕ)−1(B)= ϕ−1(g−1(B)). i ◦ i i ◦ i Hence, −1 −1 −1 µ{pi}(ϕ (B)) = piµ{pi}ϕ (gi (B)). Xi∈I 2 For the case that each fi is a contraction, the existence and uniqueness of µ{pi}i are classical and well-known (see Hutchinson [Hu81]). The existence and uniqueness of µ{pi}i for the case that each fi is a weak contraction is originally shown by [Fan96] with use of an ergodic theorem. Alternative simpler proofs are given by [AJS17, GMM, O18] 9 By this and the uniqueness of self-similar measures established in [Fan96], we have that ν = µ ϕ−1, {pi} {pi} ◦ and hence, ν (ϕ(f f (A))) µ (f f (A)). {pi} i1 ◦···◦ in ≥ {pi} i1 ◦···◦ in By the definition of µ, µ (f f (A)) = p p µ (f f )−1(f f (A)) {pi} i1 ◦· · ·◦ in j1 · · · jn {pi} j1 ◦···◦ jn i1 ◦···◦ in j ,...,j ∈I 1 Xn p p µ (A) > 0. ≥ i1 · · · in {pi} By the definition of dimH ν{pi}, dim ϕ(f f (A)) dim ν . H i1 ◦···◦ in ≥ H {pi} Therefore, dimH µ{p } + ǫ dim f f (A) α> i H i1 ◦···◦ in . dim ν ≥ dim ϕ(f f (A)) H {pi} H i1 ◦···◦ in Now the assertion follows from Lemma 3.2. 3.1. Local regularity. Let I = 0, 1,...,N 1 . { − } 3.1.1. Assumptions for (X, fi i). Hereafter, we denote the set of linear bounded transformations on{ a separable} Banach space E by B(E, E). Assumption 3.4. (i) E1 is a separable Banach space. (ii) X is a compact subset of E1 such that fi(X)= X i[∈I and its interior is non-empty. (iii) each fi is weakly contractive on X. (iv) There exists the total derivative of f at x X, which is denoted by i ∈ Dfi(x) B(E1, E1). (v) Assume∈ that for each i and x X, Df (x) is non-degenerate, specifically, ∈ i Df (x)(z) inf | i | > 0, z6=0 z | | where is the norm of E . If so, Df (x) is invertible and | · | 1 i −1 (Df (x))−1w (Df (x))−1 −1 = sup | i | k i k w w∈Dfi(x)(E1),w6=0 | | ! w Dfi(x)(z) = inf | | −1 = inf | |. w∈Dfi(x)(E ),w6=0 (Df (x)) w z6=0 z 1 | i | | | Hereafter, z denotes the maximal integer which does not larger than a real number z⌊. ⌋ Assumption 3.5 (regularity property). We say that (X, fi i) satisfies a regularity property if for any ǫ > 0 there exists C > 0 such{ that} for every 10 KAZUKIOKAMURA n 0, i ,...,i , ≥ 1 ⌊n(1+ǫ)⌋ dist f f (X), X f f (X) Cdiam f f (X) . i1 ◦···◦ i⌊n(1+ǫ)⌋ \ i1 ◦···◦ in ≥ i1 ◦···◦ i⌊n(1+ǫ)⌋ Assumption 3.6 (Existence of two distant points) . There exists c> 0 such that for every n 0 and (i ,...,i ), ≥ 1 n d (f f (fix(f )),f f (fix(f ))) c diam (f f (X)) . X i1 ◦···◦ in 0 i1 ◦···◦ in N−1 ≥ i1 ◦···◦ in Assumption 3.7 (Measure separation property). Assume that µ (f f (X) f f (X)) = 0 {pi} i1 ◦···◦ in ∩ j1 ◦···◦ jn holds whenever (i ,...,i ) = (j ,...,j ). 1 n 6 1 n 3.1.2. Assumptions for (Y, gi i). The following corresponds to Assumption 3.4. { } Assumption 3.8. (i) E2 is a separable Banach space. (ii) Y is a closed subset of E2 such that gi(Y )= Y i[∈I and its interior is non-empty. (iii) each gi is weakly contractive on Y . (iv) There exists the total derivative of g at y Y , which is denoted by i ∈ Dgi(y) B(E2, E2). (v) Assume∈ that for each i and y Y , Dg (y) is non-degenerate, specifically, ∈ i Dg (y)(z) inf | i | > 0. z∈E \{0} z 2 | | If so, Dgi(y) is invertible and −1 −1 Dgi(y)(z) (Dgi(y)) = inf | | . k k z∈E2\{0} z | | (vi) fix(g ) = fix(g ). 0 6 N−1 Y is not necessarily compact. 3.1.3. Local regularity for solution. The following gives a local regularity for ϕ at µ{pi}-almost every each point. Theorem 3.9. Let ϕ : X Y be a continuous solution of (1.1). Under Assumptions 3.4 - 3.8, we have→ the following: (i) If pi log(1/ Dgi(y) )ν{p }(dy) α< i∈I Y k k i , p log Df (x)−1 µ (dx) P i∈I iR X k i k {pi} then (α, x, 0) holds for µ -a.e. x. P{pi} R (ii) If −1 pi log Dgi(y) ν{p }(dy) β > i∈I Y k k i , p log(1/ Df (x) )µ (dx) Pi∈I i RX k i k {pi} then, (β,x, + ) holds for µ -a.e. x. ∞ P {pRi} Remark 3.10. (i) Assumptions 3.4-3.8 may not assure that there exists a solution of (1.1). 11 (ii) In the statement of [O16, Theorem 1], we had to assume that Dgi(y) is non-degenerate as in Assumption 3.8 (v), but actually we did not. (Notation here is different from [O16].) (iii) However, [O16] considers the case X = [0, 1] only. Here we give a gen- eralization for more general self-similar sets containing (d 2-dimensional) standard Sierpi´nski gaskets and carpets, for example. Ass≥umptions 3.4-3.8 hold for d-dimensional Sierpi´nski gaskets and carpets. (iv) [Ha85, Theorems 7.3 and 7.5] correspond to [Z01, Theorems 6 and 7], respectively. They are essentially same, but [Ha85, Theorems 7.3 and 7.5] is a little more general than [Z01, Theorems 6 and 7]. Proof of Theorem 3.9. For i I, let ∈ X(i) := f (X) f f (X) f f (X). i \ i1 ◦···◦ in ∩ j1 ◦···◦ jn n[≥1 (i1,...,in)[6=(j1,...,jn) e Assume that for some i, x X(i). Let T (x) := f −1(x) and I (x) := ∈ i 1 i. Then, for some i , T (x) X(i ). Hence we can define T (T (x)) = 2 ∈ 2 f −1(T (x)) and I (x) := I (T (xe)) = i . By repeating this, we have an i2 2 1 2 infinite sequence (I (x)) INefor x X(i). By Assumption 3.7, n n≥1 ∈ ∈ i∈I µ{pi}i i∈I X(i) = 1. S N e LetθSbe the one-sided shift on I . By [FL99, Proposition 1.3], there exists a measurablee map π : IN X such that → T π = π θ, on π−1 X(i) . ◦ ◦ ! i[∈I Here we put the cylindrical σ-algebra on IN. e N N For i I, let σi : I I such that σi(ω)= iω, which is a concatenation ∈ → N of i and ω. Let η{pi} be a specific probability measure on I such that η = p η σ−1. {pi} i {pi} ◦ i Xi Then θ is invariant and ergodic with respect to η{pi}. By [FL99, Proposition 1.3 (ii)], µ = η π−1. {pi} {pi} ◦ Thus we have that µ{pi} is invariant and ergodic with respect to T , and furthermore I are i.i.d. under µ . { i}i {pi} We first show assertion (ii). For x X(i), let ∈ i∈I Fn(x) := f f (fix(f0)) f f (fix(fN−1)) , I1(x) ◦···◦ In(x) S− I1(ex) ◦···◦ In(x) and, G (x) := g g (fix(g )) g g (fix(g )) . n I1(x) ◦···◦ In(x) 0 − I1(x) ◦···◦ In(x) N−1 k−1 If n = 0, let G0(x) := 1. Since Ik(x) = I1(T (x)), by [O16, Lemma 3.1], 1 F (x) + o (1) n Df (Tx) + o (1), (3.1) (Df (Tx))−1 n ≤ F (T (x)) ≤ k I1(x) k n k I1(x) k n−1 12 KAZUKIOKAMURA and 1 G (x) + o (1) n Dg (ϕ(Tx)) + o (1), (Dg (ϕ(Tx)))−1 n ≤ G (T (x)) ≤ k I1(x) k n k I1(x) k n−1 (3.2) where the small order on(1) is uniform with respect to x. For every ǫ> 0, there exists N such that for every n > N, n n−i Gi(T (x)) log Gn(x)= log n−i+1 Gi−1(T (x)) Xi=1 n−N n−i+1 −1 log 1/ Dg n−i (ϕ(T (x))) log(1 + ǫ), ≥ k I1(T (x)) k − Xi=1 and, n−N n−i Fi(T (x)) logFn(x)= CN + log n−i+1 Fi−1(T (x)) Xi=1 n−N n−i+1 CN + Df n−i (T (x)) + log(1 + ǫ), ≤ k I1(T (x)) k Xi=1 where C is a constant independent from ǫ. Now by using these inequalities and the Birkhoff ergodic theorem it holds that µ{pi}-a.e.x, log Gn(x) −1 lim inf log 1/ DgI (x)(ϕ(Tx)) µ{p }(dx) n→∞ n ≥ k 1 k i ZX = p log 1/ Dg (ϕ(Tx))−1 µ (dx) i k i k {pi} i ZX X = p log 1/ Dg (ϕ(x))−1 µ (dx) i k i k {pi} i ZX X −1 = pi log 1/ Dgi(y) ν{pi}(dy), Y k k Xi Z and, log F (x) lim sup n log Df (Tx) µ (dx) n ≤ k I1(x) k {pi} n→∞ ZX = pi log Dfi(x) µ{pi}(dx) X k k Xi Z 1 −1 pi log 1/ Dgi(y) ν{pi}(dy). ≤ β Y k k Xi Z Hence, log Gn(x) lim sup β, µ{pi}-a.e.x. n→∞ log Fn(x) ≤ Essentially the same argument as above is done in the proof of [O16, Theo- rem 1.1]. Let ǫ> 0. Then, we have that for µ{pi}-a.e.x, it holds that for sufficiently large n, Fn(x) < 1 and Gn(x) < 1, and hence, G (x) F (x)β+ǫ. n ≥ n 13 Let F (x) := f f (fix(f )) x . 0,n I1(x) ◦···◦ In(x) 0 − FN−1,n(x) := x fI1(x) fIn(x)(fix(fN− 1)) . − ◦···◦ G0,n(x) := g I1(x) gIn(x)(fix(g0)) ϕ(x) ◦···◦ − = ϕ(fI1( x) fIn(x)(fix(f0))) ϕ(x) . ◦···◦ − GN−1,n( x) := ϕ(x) gI1(x) gIn(x)(fix(g N−1)) − ◦···◦ = ϕ(fI1(x) fIn(x)(fix(fN−1))) ϕ(x) . ◦···◦ − Then, by Assumption 3.6, 2max G (x), G (x) G (x) (c/2)β+ǫ(F (x)+ F (x))β+ǫ. { 0,n N−1,n }≥ n ≥ 0,n N−1,n Hence, there exists a constant c> 0 such that G (x) G (x) max lim sup 0,n , lim sup N−1,n c, µ -a.e.x. F (x)β+ǫ e F (x)β+ǫ ≥ {pi} n→∞ 0,n n→∞ N−1,n This completes the proof of assertion (ii). e We second show assertion (i). Let n(x,x′) := min k : I (x) = I (x′) , x,x′ X(i). k 6 k ∈ i[∈I ′ We compare N −n(x,x ) with x x′ . By the definition ofe I , | − | { i}i x f f (X), ∈ I1(x) ◦···◦ I⌊n(x,x′)(1+ǫ)⌋(x) and x′ X f f (X). ∈ \ I1(x) ◦···◦ In(x,x′)(x) Let ǫ> 0. Then by Assumption 3.5, x x′ c diam f f (X) . | − |≥ I1(x) ◦···◦ I⌊n(x,x′)(1+ǫ)⌋(x) Now we give an upper bound for ϕ(x) ϕ(x′) . Due to (1.1), both of the ϕ(x) and ϕ(x′) are contained in | − | ϕ f f (X) = g g (ϕ(X)) I1(x) ◦···◦ In(x,x′)−1(x) I1(x) ◦···◦ In(x,x′)−1(x) g g (Y ). ⊂ I1(x) ◦···◦ In(x,x′)−1(x) Therefore, ϕ(x) ϕ(x′) diam g g (Y ) . | − |≤ I1(x) ◦···◦ In(x,x′)−1(x) By noting (3.1), (3.2), µ{pi}i i∈I X(i) = 1, and the fact that ϕ is continuous, it suffices to show that S e diam g g (Y ) I1(x) ◦···◦ I⌊n(1+ǫ)⌋(x) lim inf α = 0, µ{p }-a.s.x. n→∞ diam f f (X) i I1(x) ◦···◦ In−1(x) By [O16, Lemma 3.1], we have that 1 diam fI1(x) fIn(x)(X) +on(1) ◦···◦ Df (Tx) +on(1), (Df (Tx))−1 ≤ diam f f (X) ≤ k I1(x) k k I1(x) k I1(x) ◦···◦ In−1(x) 14 KAZUKIOKAMURA and 1 diam gI1(x) gIn(x)(Y ) +on(1) ◦···◦ Dg (ϕ(Tx)) +on(1), (Dg (ϕ(Tx)))−1 ≤ diam g g (Y ) ≤ k I1(x) k k I1(x) k I1(x) ◦···◦ In−1(x) which correspond to (3.1) and (3.2), respectively. In the same manner as in the proof of assertion (i), we have that log diam gI1(x) gI⌊n(1+ǫ)⌋(x)(Y ) lim sup ◦···◦ n→∞ n (1 + ǫ) pi log Dgi(y) ν{pi}(dy) , µ{pi}-a.s.x. ≤ Y k k ! Xi Z and, log diam fI1(x) fIn−1(x)(X) −1 lim inf ◦···◦ (1 ǫ) pi log 1/ Dfi(y) µ{pi}(dy) n→∞ n ≥ − X k k ! Xi Z 1 2ǫ − pi log Dgi(y) ν{pi}(dy) , µ{pi}-a.s.x, ≥ α(1 ǫ/2) Y k k ! − Xi Z where we have used the assumption of (i). We remark that if we take sufficiently small ǫ> 0, 1 2ǫ − < 1+ ǫ. 1 ǫ/2 − By using this and pi log Dgi(y) ν{pi}(dy) < 0, Y k k Xi Z we have assertion (i). 4. Examples for existence and uniqueness 4.1. Examples for the case that gi does not depend on x. Example 4.1 (The topological structure of X is not given). (i) Assume that I contains at least two distinct points i0, i1 . If fi : Xi X is an identity map for each i I, then, A = X {X .} If X X →= and Y ∈ i0 ∩ i1 i0 ∩ i1 6 ∅ contains at least two distinct points y0,y1 and gik yk, k = 0, 1, then, Assumption 2.2 fails and hence there{ is no} solutions for≡ (1.1). This is also an example such that the compatibility conditions in [SB17, Definition 1] fails. (ii) In (i) above, it is crucial to assume that X X = . Indeed, if 0 ∩ 1 6 ∅ I = 0, 1 , X0 = S X and X1 = X S, S and X S are both non-empty, f are{ identity} maps,⊂ Y contains at least\ two distinct\ points y ,y , and i { 0 1} gk yk, k = 0, 1, then, a function ϕ : X Y such that ϕ = y0 on S and ϕ =≡y on X S gives a solution for (1.1).→ 1 \ We now give examples for X = Y = [0, 1]. Here and henceforth, we always give [0, 1] the topology induced by the Euclid metric. First we consider the case that A = 0, 1 . { } 15 Proposition 4.2 (Existence and continuity). Let X = Y = [0, 1]. As- sume that each fi is a weak contraction in the sense of Matkowski on [0, 1] satisfying that f (0) = 0, f (1) = 1, f (1) = f (0), 1 i N 1. 0 N−1 i−1 i ≤ ≤ − f (0) f (x) f (1), x [0, 1]. i ≤ i ≤ i ∈ f (x) > 0,f (x) < 1, x (0, 1). 0 N−1 ∈ Assume that each gi is a strictly increasing weak contraction in the sense of Matkowski on [0, 1] such that g (0) = 0, g (1) = 1, g (1) = g (0), 1 i N 1. 0 N−1 i−1 i ≤ ≤ − Then, the unique solution of (1.1) exists and is continuous. Remark 4.3. We do not assume that each fi is injective. Therefore, the method taken in [R10] is not applicable to this case, however, in the following proof we heavily depend on the order structure of X = [0, 1]. Proof. Assumptions 2.1, 2.2 and 2.3 hold. Hence by Theorem 2.9, the unique solution of (1.1) exists. Now we show the continuity of the solution. First we recall the following simple assertion. Lemma 4.4. Let D [0, 1] be a dense subset of [0, 1] and φ : D [0, 1] is increasing. Let ⊂ → D D φ+ (x) := lim φ(y), x [0, 1), φ+ (1) := 1. y→x,y∈D∩(x,1] ∈ D D φ− (x) := lim φ(y), x (0, 1], φ− (0) := 0. y→x,y∈D∩[0,x) ∈ D D Then, φ+ and φ− are right and left continuous, respectively. It holds that A is dense in [0, 1] and ϕ is increasing on A. Since all f , g , 0 i N 1, are continuous, the left and right continuous i i ≤ ≤ − modifications ofe the solution of (1.1) is also the solution ofe(1.1). Hence, by the uniqueness of (1.1), these functions are identical with each other. By this and Lemma 4.4, the unique solution is left and right continuous, and hence, continuous. In the following example, we deal with the Minkowski question-mark function, which is often denoted by ?(x). It was introduced by Minkowski [Min1904] in 1904 and has been closely connected with continued fractions, number theory and dynamical systems. It is defined by ∞ ?(x)= ( 1)k−121−(n1+···+nk), − Xk=1 if x [0, 1] has the following continued fraction expansion: ∈ 1 x = [n1,n2,... ]= . n + 1 1 n2+··· There is another way of (equivalent) definition which uses the Farey se- quence. De Rham [dR57] regarded ?(x) as a solution of a functional equa- tion. Singularity of the Minkowski question-mark function was considered by Denjoy [D38] and Salem [Sa43]. Some analytic properties of the Minkowski 16 KAZUKIOKAMURA question-mark function were considered by Viader, Paradis and Bibiloni [VPB98], Kesseb¨ohmer and Stratmann [KS08], Jordan and Sahlsten [JS13], etc. The inverse function of the Minkowski question-mark function is called the Conway box function. Analytic properties of the inverse were recently considered by Mantica and Totik [MT18+]. Example 4.5 (Equation driven by weak contractions). Let I = 0, 1 , { } X = Y = [0, 1]. Let f0(x)= x/2, f1(x) = (x + 1)/2, g0(y)= y/(y + 1) and g1(y) = 1/(2 y). Neither g0 or g1 is a contraction, but both of them are weak contractions− in the sense of Browder. Then, (i) A unique solution ϕ of (1.1) is the Conway box function, that is, the inverse function of the Minkowski question-mark function. (ii) For any dyadic rational x on [0, 1] and a> 0, f(y) f(x) f(x) f(y) lim sup − a = limsup − a =+ . (4.1) y→x,y>x x y y→x,y This leads a contradiction. (ii) Let a = 1/2 and b = 3/4. Then, A = 0, 1 and the compatibility condition of [SB17, Definition 1] fails. { } (iii) Let a = 2/3 and b = 3/4. Assume that ϕ is a solution for (1.1). Then, there exist two candidates for the value of ϕ(1/2). This leads a contradiction. 4.2. Examples for the case that gi depends on x. Example 4.8 (A case that a unique solution of (1.1) is not continuous). Fix i I. Assume that X is a topological space and fi is continuous. Assume∈ that there exists a continuous function G : Y Y Y such that G : X Y Y does not depend on y and is not continuous× → as a function i × → on X, where we let Gi(x,y)= G(gi(x,y),y). Then, under Assumptions 2.1, 2.2, 2.3 and 2.4, a unique solution of (1.1) is not continuous. For example, this is applicable to the case that X = [0, 1], f0(x) = x/2, f (x) = (x + 1)/2, Y = R, g (x) = px, g (x) = (1 p)x + p, p (0, 1), h : 1 0 1 − ∈ X Y is a non-continuous function, gi(x,y)= h(x)+gi(y) and G(y1,y2) := y →g (y ). 1 − i 2 e e Example 4.9 (Sierpi´nski gasket). Let V = q ,q ,qe R2 be a set of 0 { 0 1 2} ⊂ threee points such that qi qj = 1, i = j. Let fi(x) = (x + qi)/2, i = 0, 1, 2, 2 k − k 6 2 x R . Let X be a unique compact subset of R such that X = i∈I fi(X). For∈ i I and x X, let g (x, ) be weak contractions on R such∪ that ∈ ∈ i · gi(qj, fix(gj)) = gj(qi, fix(gi)) for each (i, j). (4.2) A = V0 and Assumptions 2.1, 2.2, 2.3 and 2.4 hold. Therefore by Theo- rem 2.5, there exists a unique solution of (1.1). This case is considered by [CKO08], [R10] and [RR11]. We remark that by (4.2), if each gi( )= gi(x, ) is linear, then, the solution is quite limited. · · 4.3. Example for the case that infinitely many solutions exist. In this subsection, we consider the case that at least one gi is not weakly contractive, that is, Assumptions 2.1 and 2.2 hold, but Assumption 2.3 fails. Before we delve into the main result, we give an instructive example. Example 4.10 (Example for the case that the uniqueness fails; Cf. [O14T]). Consider (1.1) for the case that I = 0, 1 , X = Y = [0, 1],f (x) = (x + { } i i)/2, gi(y)=Φ(Au,i; y), i = 0, 1, where we let az + b a b Φ(A; z)= for A = , and, cz + d c d xu 0 0 xu 2 Au,0 = , Au,1 = , xu = . u2x2 1 u2x2 1 u2x2 √ 2 − u − u − u 1+ 1 + 8u In this case, by [O14T, Proposition 3.4], it holds that for any dyadic rational x, lim ϕu(y) < ϕu(x). y→x,y Theorem 4.11. Let X = Y = [0, 1] and fi(x) = (x + i)/2, i = 0, 1. Then there exist strictly increasing functions g0 and g1 on [0, 1] such that As- sumption 2.4 holds and furthermore there exist infinitely many solutions for (1.1). Proof. Recall that M is the inverse function of Minkowski’s question-mark function appearing in Example 4.5. Let g0(x) := M(x)/2 and g1,0(x) := (x + 1)/2. We define strictly increasing functions g : [0, 1] [0, 1], n 1, 1,n → ≥ in the following manner. We have that g1,0 g0(0) = 1/2 and g1,0 g0(1) = 3/4. By the intermediate value theorem,◦ there exists at least o◦ne point y (1/2, 3/4) such that g g (y ) = x . Then there exists a strictly 0 ∈ 1,0 ◦ 0 0 0 increasing continuous function g1,1 such that sup g1,1(x) g1,0(x) 1/8, x∈[0,1] | − |≤ min y 1/2, 3/4 y g (x)= g (x), x y > { 0 − − 0}. 1,1 1,0 | − 0| 8 and, for a dyadic rational x1 (1/2, 3/4), g1,1 g0(x1)= x1. This is possible because we can take a sequence∈ z of dyadic◦ rationals converging to y , { n}n 0 and due to the continuity of g0, g0(zn) g0(y0) as n . Now by (4.1), there exists a point y→> x such that→∞g g (y ) = y . 2 1 1,1 ◦ 0 2 2 Then there exists a strictly increasing continuous function g1,2 such that sup g1,2(x) g1,1(x) 1/16, x∈[0,1] | − |≤ min y x , 3/4 y g (x)= g (x), x y > { 2 − 1 − 2}. 1,2 1,1 | − 2| 16 and, there exists at least one dyadic rational x (x , 3/4) such that g 2 ∈ 1 1,2 ◦ g0(x2) = x2. By repeating this procedure, we can define the limit g1 := lim g . Then, g is an increasing continuous function such that g n→∞ 1,n 1 1 ◦ g0(xn)= xn for any n 1, and, g1(0) = 1/2 and g1(1) = 1. We remark that ≥ C := f f ( 1/3, 2/3 ) i1 ◦···◦ in { } n[≥1 i1,...,i[n∈I satisfies that f −1(C) C, i ⊂ [i so the values of each solution of (1.1) on the set C is determined indepen- dently on the value of ϕ on X C. If we give a value of ϕ(1/3), then, all the values of ϕ on C are uniquely\ determined by (1.1). Since 1/3 is a fixed point of f f on [0, 1], 0 ◦ 1 ϕ(1/3) = ϕ(f f (1/3)) = g g (ϕ(1/3)). 0 ◦ 1 0 ◦ 1 Hence ϕ(1/3) is a fixed point of g g . By our choice of g and g , there 0 ◦ 1 0 1 exist at least countably many fixed points of g0 g1 in [0, 1], and hence there are at least countably many candidates of ϕ(1◦/3), and each value can be taken. 19 It will be easy to see that Assumption 2.4 holds. Remark 4.12. The idea of introducing C in the proof is similar to that in introducing Assumption 2.4. 5. Examples for regularity 5.1. The case that X = [0, 1]. Example 5.1 (linear case). Let X = Y = [0, 1]. Let f0(x) = x/2 and f1(x) = (x + 1)/2. Let a (0, 1). Let g0(y)= ay and g1(y) = (1 a)y + a. Then, the solution ϕ of (1.1)∈ is Legesgue’s singular function. (We− remark that by a direct use of (1.1), it holds that ϕ is ( log max a, 1 a / log 2)- H¨older continuous.) Let 0 (Such g0 and g1 exist e for any ǫ> 0.) We will e show that under this setting, the solution (1.1) has derivative zero at every dyadic rational. Proof. Let 0 We remark that νp, 1−p depend on the choice of (p, g0, g1) so hereafter we write the measure as νp,g0,g1 . p log (g )′(y) (1 p) log (g )′(y) ν (dy) − 0 − − 1 p,g0,g1 Z[0,1] e e 20 KAZUKIOKAMURA = (pν ([0, 1/2]) + 2(1 p)) log 2 p,g0,g1 − (use g (1) > g (1) = 3/4 and g g (1) > g g (1).) 0 0 0 ◦ 0 0 ◦ 0 3 (pνp,g0,g1 (g0 g0([0, 1])) + 2(1 p)) log 2 = (p 2p + 2) log 2. ≤ e ◦ − e e − If we let p = 3/4, then, p3 2p + 2 < 1. Hence if we take sufficiently small ǫ> 0, − p log g′ (y) (1 p) log g′ (y) ν (dy) < log 2. − 0 − − 1 p,1−p Z[0,1] In some cases, computation for ′ pi log(1/gi(y))ν{pi}(dy) [0,1] Xi∈I Z is hard, because gi can be a non-linear function and hence the integral depends essentially on ν{pi}. Lemma 5.3. Let (Z, dZ ) be a compact metric space and hi i be a collection of weak contractions in the sense of Browder on Z satisfying{ } that Z = hi(Z). i[∈I Assume that η = p η h−1. {pi} i {pi} ◦ i Xi∈I Then, for each fixed i, as p 1, η converges weakly to the delta measure i → {pi} on fix(hi). Proof. Let F be a real continuous bounded function on Z. Let ǫ> 0. Then, by using the fact that hi is a weak contraction, we can take a sufficiently large n such that n sup F hi (z) F (fix(hi)) ǫ. z∈Z | ◦ − |≤ By the definition of η{pi}, F dη pn F hndη (1 pn)sup F (z) . {pi} − i ◦ i {pi} ≤ − i | | ZZ ZZ z∈Z Hence for every pi sufficiently close to 1, F dη F (fix(h )) 3ǫ. {pi} − i ≤ ZZ Proposition 5.4. Let I = 0, 1,...,N 1 . Let X = Y = [0, 1]. Assume 1 { − } that fi and gi are C functions on [0, 1] such that 0= f0(0) = g0(0) fi−1(1) = fi(0), gi−1(1) = gi(0), 1 i N 1, ′ ′ ≤ ≤ ′ − ′ 0 < min inf min fi (z), gi(z) max sup max fi (z), gi(z) < 1. i∈I z∈[0,1] { }≤ i∈I z∈[0,1] { } ′ ′ Fix i I. Let α> log (1/ gi(fix(gi)) ) / log(1/ fi (fix(fi)) ). Then, (α, U, + ) holds∈ for every open set U| . | | | ∞ 21 Proof. By Lemma 5.3, ′ ′ i∈I pi log(1/ gi(y) )ν{pi}(dy) log 1/ gi(fix(gi)) lim | ′ | = | ′ | < α. pi→1 p log 1/ f (x) µ (dx) log 1/ f (fix(f )) P i∈I iR | i | {pi} | i i | Now use Theorem 3.9 for an appropriate (p ,...p ). Then, (α, x, + ) P R 0 N−1 ∞ holds for µ{pi}-a.e.x. We remark that µ{pi}(U) > 0, and then the assertion follows. Proposition 5.5. Let I = 0, 1,...,N 1 . Let X = Y = [0, 1]. Assume { − } that fi, gi are strictly increasing continuous functions on [0, 1] such that 0= f0(0) = g0(0) Remark 5.7 (Formula for µϕ). We assume that the solution ϕ satisfies 2 the Dini condition in [FL99] and gi C ([0, 1]) holds for each i. By the Lebesgue-Stieltjes integral, for every∈ bounded Borel measurable function 3A singular function is a continuous increasing function on the unit interval whose deriva- tives are zero at Lebesgue almost surely points. 22 KAZUKIOKAMURA F : [0, 1] R, → ′ F (x)dµϕ(x)= gi(ϕ(x))F (fi(x))dµϕ(x). [0,1] [0,1] Z Xi Z Let h be a positive function on [0, 1] such that ′ h(x)= gi(gi(ϕ(x)))h(fi(x)). Xi h ϕ−1(x)= g′(g (x))h ϕ−1(g (x)). ◦ i i ◦ i Xi This is unique under the constraint that h(x)µϕ(dx) = 1. Z[0,1] See [FL99, Theorem 1.1]. Then, by [FL99, Corollary 3.5], h(x) log g′(ϕ(x))µ (dx) i [fi(0),fi(1)] i ϕ dimH µϕ = h(x) log f ′(f −1(x))µ (dx) Pi [Rfi(0),fi(1)] i i ϕ Let fi(x) = (x + i)/NP, i R I. Then, ∈ ′ ′ i∈I [0,1] H(gi(y))gi(y) log (1/gi(gi(y))) ℓ(dy) dimH µϕ = P R log N where ℓ is the Lebesgue measure on [0, 1], ′ H(y)= gi(gi(y))H(gi(y)), and H(y)ℓ(dy) = 1. [0,1] Xi∈I Z It is interesting to investigate properties for H. 5.2. The case that X is the two-dimensional Sierpi´nski gasket. Fi- nally we deal with an example for the case that X is the two-dimensional Sierpi´nski gasket and Y = [0, 1]. Example 5.8. Assume that X is the two-dimensional Sierpi´nski gasket and Y = [0, 1]. Here we follow the notation in Example 4.9, however, we consider the case that each g does not depend on x. Let g (y) = 1/(2 y) 1/2, i 0 − − g1(y) = (2/3)y + 1/6 and g2(y)= y/(y + 1) + 1/2. Then, by Theorem 2.5, the unique solution ϕ of (1.1) holds, and by adopt- ing the method4 taken in [R10], we can show that ϕ is continuous. Furthermore, by Theorem 3.9, if 1 3 2 2 3 [0,1] log 2 (2 y + y ) dν(1/3,1/3,1/3)(y) β > − , R log 2 then, (β,x, + ) holds for µ -a.e. x. ∞ (1/3,1/3,1/3) Remark 5.9. [CKO08, R10, RR11] consider fractal interpolation functions on Sierpi´nski gasket and more generally post-critically finite self-similar sets. Our choices of g are not considered by them, and it seems that their { i}i 4Recall Remark 4.3 23 methods are not applicable to showing that (α, x, + ) holds for µ(1/3,1/3,1/3)- a.e. x. ∞ Remark 5.10 (Stability). By arguing as in [O16, Proposition 2.7 (iii)], we are able to extend Proposition 5.4 to several cases of non-differentiability of gi. The regularity assumptions for fi and gi are not necessarily essential. 6. Stability Definition 6.1. Let X and Y be two compact metric spaces. We let the Gromov-Hausdorff distance dGH(X,Y ) be the infimum of any values Haus dM (f(X), g(Y )) for any compact metric space M and any isometric em- Haus beddings f : X M and g : Y M, where dM is the Hausdorff metric on the space of all→ compact subsets→ of the space M. Let I be a finite set containing at least two points. Let n N . Let ∈ ∪ {∞} Z(n) be a compact metric space. For i I and n N , let h(n) be ∈ ∈ ∪ {∞} j a weak contraction on Z(n) in the sense of Browder [B68]. Let K(n) be the (n) (n) (n) attracter of the iterated function system Z , (hi )i∈I , that is, K is (n) (n) (n) (n) the unique compact subset of Z satisfying that K = i∈I hi K . (n) (n) (n) S Theorem 6.2. Let Ki , K be subsets of Z satisfying that K(n) = h(n) K(n) , (6.1) e e i i i[∈I and e e Haus (n) (n) lim max d (n) Ki , K = 0. (6.2) n→∞ i∈I Z (n) Assume that there exist compact metric spaces (M , d (n) ) and e e M n∈N∪{∞} (k) (n) isometries ϕn,k : Z M , k n, such that for each i I, (a) → ∈ { ∞} ∈ Haus (n) (n) (∞) lim d (n) ϕn,n(K ), ϕn,n(K ) ϕn,∞(Z ) = 0, n→∞ M i i ∩ and, (b) e e (n) −1 (∞) −1 lim sup d (n) ϕn,n h ϕ (x), ϕn,∞ h ϕ (x) = 0, n→∞ M i n,n i n,∞ x∈L(n) ◦ ◦ ◦ ◦ where we let L(n) := ϕ (K(n)) ϕ (Z(∞)). j n,n j ∩ n,∞ Then, Haus e (n) (∞) lim d (n) ϕn,n(K ), ϕn,∞(K ) = 0. n→∞ M In particular e lim dGH K(n), K(∞) = 0. n→∞ (n) (n)e (n) (n) (n) (n) If K is the attracter of Z , (hi )i∈I , then, Ki = K = K satisfy the assumption. { } e e 24 KAZUKIOKAMURA Proof. We show this assertion by contradiction. Assume the conclusion fails. Then, by relabeling if needed, there exists ǫ0 > 0 such that for any n, dHaus ϕ (K(n)), ϕ (K(∞)) ǫ . M (n) n,n n,∞ ≥ 0 By [J97, Theorem 1 (e)], for each i I, we can take an upper semicontin- e ∈ (∞) uous function φi such that φi(t) inf F (t) ǫ1 > 0. (6.3) t≥ǫ0 ≥ It follows that for each j I, ∈ Haus (n) (n) (∞) (∞) dM (n) ϕn,n hj (Kj ) , ϕn,∞(hj (K )) Haus (n) (n) (n) −1 (n) d (n) ϕn,n(h (Ke )), ϕn,n h ϕ L ≤ M j j ◦ j ◦ n,n j +dHaus ϕ h(n) ϕ−1 L(n) , ϕ h(∞) ϕ−1 L(n) M (n) n,n ◦ j ◦ n,ne j n,∞ ◦ j ◦ n,∞ j (∞) (n) (∞) +dHaus ϕ h ϕ−1 L , ϕ h ϕ−1 ϕ (K(∞)) . M (n) n,∞ ◦ j ◦ n,∞ j n,∞ ◦ j ◦ n,∞ n,∞ (n) (n) (1) By using the facts that K is an attracter and each hj is weakly contractive and the assumption (a), dHaus ϕ (K(n)), ϕ h(n) ϕ−1 L(n) dHaus ϕ (K(n)),L(n) 0. M (n) n,n n,n ◦ j ◦ n,n j ≤ M (n) n,n j j → (2) By the assumption (b), e dHaus ϕ h(n) ϕ−1 L(n) , ϕ h(∞) ϕ−1 L(n) 0. M (n) n,n ◦ j ◦ n,n j n,∞ ◦ j ◦ n,∞ j → (∞) (3) By using that facts that each hj is a weak contraction and (6.3), dHaus ϕ h(∞) ϕ−1 L(n) , ϕ h(∞) ϕ−1 ϕ (K(∞)) M (n) n,∞ ◦ j ◦ n,∞ j n,∞ ◦ j ◦ n,∞ n,∞ φ dHaus L(n), ϕ (K(∞)) . ≤ j M (n) j n,∞ (n) (n) (n) φ dHaus ϕ (K ), ϕ (K(∞)) + dHaus L , ϕ (K ) . ≤ j M (n) n,n j n,∞ M (n) j n,n j Since for eachj, e e Haus (n) (n) lim d (n) ϕn,n(K ), ϕn,n(K ) = 0, n→∞ M j by recalling that e e Haus (n) (∞) dM (n) ϕn,n(K ), ϕn,∞(K ) >ǫ0, it follows that for large n, e Haus (n) (∞) dM (n) ϕn,n(Kj ), ϕn,∞(K ) >ǫ0. Hence, for large n, e Haus (n) (∞) Haus (n) (n) φj dM (n) ϕn,n(Kj ), ϕn,∞(K ) + dM (n) Lj , ϕn,n(Kj ) e e 25 dHaus ϕ (K(n)), ϕ (K(∞)) + dHaus L(n), ϕ (K(n)) ǫ ≤ M (n) n,n j n,∞ M (n) j n,n j − 1 (n) dHaus ϕ (K(n)), ϕ (K(∞)) + dHaus ϕ (K ), ϕ (K(n)) ≤ M (n) n,n e n,∞ M (n) n,n j en,n (n) (n) +dHaus L , ϕ (K ) ǫ e M (n) j n,n j − 1 e e dHaus ϕ (K(n)), ϕ (K(∞)) ǫ /2. ≤ M (n) n,n n,∞e − 1 Hence if n is sufficiently large, e Haus (n) (∞) dM (n) ϕn,n(K ), ϕn,∞(K ) Haus (n) (n) (∞) (∞) max d (n) ϕn,n h e(K ) , ϕn,∞(h (K )) ≤ j M j j j Haus (n) (∞) d (n) ϕn,n(K )e, ϕn,∞(K ) ǫ1/4. ≤ M − This is a contradiction. e (n) (n) 6.1. Application. For n N , let (X , d (n) ) and (Y , d (n) ) be ∈ ∪ {∞} X Y two compact metric spaces. Let f (n) : X(n) X(n) and g(n) : Y (n) Y (n) j → j → be weak contractions in the sense of Browder. Consider a conjugate equation (n) (n) (n) (n) (1.1) on X ,Y , (fi )i∈I , (gi )i∈I satisfying Assumptions 2.1 - 2.2. Let (n) (n) (n) hi (x,y) := fi (x), gi (y) . Let Z(n) := X(n) Y (n) and × (n) (n) (n) (n) (n) (n) 2 (n) (n) 2 dZ(n) (x1 ,y1 ), (x2 ,y2 ) := dX(n) x1 ,x2 + dY (n) y1 ,y2 . r This gives a metric on Z and by this metric Z is a compact metric space. The following is easy to see. Proposition 6.3. Let fn,n 1, and f be a uniformly continuous family of real functions on a common≥ compact metric space. Then, f f,n n → → ∞ uniformly if and only if the graphs of fn converges to the graph of f with respect to the Hausdorff distance. (n) (n) Remark 6.4. If the value of gi (x,y) depends on x, then, hi may not be a weak contraction on X Y and the proof of Theorem 6.2 does not applicable to that case. × 6.2. Examples. Now we give three cases. We consider the case that Z(∞) = M (n) only. (n) (∞) (n) Example 6.5 (Case 1, Z = Z = M and both of ϕn,n and ϕn,∞ are the identity map.). This is the case that the spaces are common and functions driving (1.1) vary. [O16, Proposition 2.7] states a result of this kind, and Theorem 6.2 will imply [O16, Proposition 2.7]. (n) (∞) (n) Example 6.6 (Case 2, Z Z = M and both of ϕn,n and ϕn,∞ are the inclusion maps.). This gives⊂ a discrete approximation of solution. Since (n) each hi is weakly contractive, it follows that K(∞) = h(∞) h(∞) fix(h(∞)) . i1 ◦···◦ ik−1 ik k[≥1 i1,...,i[k∈I 26 KAZUKIOKAMURA See [Ha85]. Therefore if we let K(n) := h(∞) h(∞) fix(h(∞)) , i1 ◦···◦ in−1 in i1,...,i[n∈I and, K(n) := K(n−1), i I, i ∈ then, (6.1) and (6.2) hold. (n) (∞) (n) Example 6.7 (Case 3, Z = Z = M and ϕn,∞ are identity maps.). We give an example for small6 deformations of de Rham type functions. Let I := 0, 1 . For n N, X(n) := [ 1/n, (n + 1)/n], Y (n) := [1/n, (n 1)/n]. (∞){ }(∞) ∈ − (n) (n) − X = Y := [0, 1]. Let eX,n : X [0, 1] and eY,n : Y [0, 1] be affine maps such that → → e ( 1/n) = 0, e ((n + 1)/n) = 1, e (1/n) = 0, e ((n 1)/n) = 1. X,n − X,n Y,n Y,n − Let f0(x)= x/2, f1(x) = (x + 1)/2, g0(y)= y/3 and g1(y) = (2y + 1)/3. Then by Proposition 4.2, there exists a unique continuous increasing solution ϕ(∞) of (1.1) for X(∞), f (∞) ,Y (∞), g(∞) . Let f (n) := e−1 f { i }i∈I { i }i∈I i X,n ◦ i ◦ e , and g(n) := e−1 g e , i I. Then there exists a unique continuous X,n i Y,n ◦ i ◦ Y,n ∈ increasing solution ϕ(n) of (1.1) for X(n), f (n) ,Y (n), g(n) . { i }i∈I { i }i∈I (n) 2 Let M := [0, 1] . Let ϕn,n(x,y) := (eX,n(x), eY,n(y)) and ϕn,∞(x,y) := (x,y). Now by applying Theorem 6.2, dGH Graph(ϕ(n)), Graph(ϕ(∞)) 0, n . → →∞ 7. Open problems (1) As in Example 4.7, let I = 0, 1 and X = Y = [0, 1]. Assume that { } f0(x)= ax, f1(x)= ax + 1 a, g0(x)= bx and g1(x)= bx + 1 b for some a, b [1/2, 1). Then, find a− pair (a, b) such that the solution of− (1.1) exists. The∈ case that a is not an algebraic number seems interesting. (2) Several notions of bounded variation functions on metric measure spaces are proposed by Miranda [Mir03], Ambrosio-Di Marino [AD14] etc. It is interesting to consider whether the solution of (1.1) is of bounded variation in the senses of their papers. Furthermore, in the case that the solution of (1.1) is not of bounded variation, it is also interesting to ask whether there exist a metric and a measure on X such that the solution of (1.1) is of bounded variation. (3) Muramoto and Sekiguchi [MS] considered a generalization of de Rham type equations on [0, 1] which is different from ours. They introduced a new class of Takagi function by using it. (4) By following Hata-Yamaguti [HY84], derivative of the solution of de Rham function with respect to a parameter in gi i yields a fractal function. From this viewpoint, it seems to be able to define{ } an analogue of Takagi function on Sierpi´nski gasket. 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